\(\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(x+1\right)\left(4-x\right)}=5\) ( ĐK : \(-1\le x\le4\) )
Đặt \(\sqrt{x+1}+\sqrt{4-x}=a\Rightarrow a^2=5+2\sqrt{\left(x+1\right)\left(4-x\right)}\)
\(\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{a^2-5}{2}\)
Phương trình trở thành :
\(a+\frac{a^2-5}{2}=5\)
\(\Leftrightarrow a^2+2a-15=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-3=0\\a+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3\left(N\right)\\a=-5\left(L\right)\end{matrix}\right.\)
\(\Leftrightarrow5+2\sqrt{\left(x+1\right)\left(4-x\right)}=9\)
\(\Leftrightarrow\left(x+1\right)\left(4-x\right)=4\)
\(\Leftrightarrow-x^2+3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\left(TM\right)\)